Spectral radius of a block circulant matrix

Question

Let $A$ the block matrix given by the blocks: $$\tilde{A}=\begin{pmatrix} 0&-\mu&0&...&0&-\mu\\ -\mu&0&-\mu&...&0&0\\ 0&-\mu&0&...&0&0\\ ...&...&...&...&...\\ 0&0&0&...&0&-\mu\\ -\mu&0&0&...&-\mu& \end{pmatrix}\quad\quad A^*=\mathbf{diag}(-\mu,-\mu,...,-\mu) $$ So $$ A=\begin{pmatrix} \tilde{A}&A^*&0&...&0&A^*\\ A^*&\tilde{A}&A^*&0&...&0\\ 0&A^*&\tilde{A}&...&...&0\\ ...&...&...&...&...&...\\ A^*&0&...&0&A^*&\tilde{A} \end{pmatrix} $$ Does anyone know some trick to calculate spectral radius of $A$? Maybe I can use the matrix norm defined by $||A||_1 = \max_{1\le j \le n} \sum_{i=1}^n |a_{ij}|$

Answer 1

By Gerschgorin disc theorem, $\rho(A)\le4|\mu|$. Since $Ae=-4\mu e$ (where $e$ is the vector of ones), we conclude that $\rho(A)=4|\mu|$.

Actually we can obtain all eigenvalues of $A$ explicitly. Let $S_r$ be the real symmetric $r\times r$ matrix $$ \pmatrix{0&1&0&\cdots&0&1\\ 1&0&1&\ddots&\vdots&0\\ 0&1&\ddots&\ddots&0&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&0\\ 0&\cdots&0&\ddots&0&1\\ 1&0&\cdots&0&1&0}. $$ This matrix is circulant and its eigenvalues admit an explicit formula: $$ \lambda_k^{(r)}=e^{2\pi ik/r}+e^{2\pi ik(r-1)/r}=2\cos\left(\frac{2k\pi}{r}\right),\ k=0,1,\ldots,r-1. $$ Now suppose that $\tilde{A}$ (or $A^\ast$) has $m$ rows and $A$ has $n$ block rows. Then $A=-\mu(I_n\otimes S_m+S_n\otimes I_m)$. Therefore the eigenvalues of $A$ are given by $$ -\mu(\lambda_k^{(m)}+\lambda_l^{(n)}) =-2\mu\left(\cos\left(\frac{2k\pi}{m}\right)+\cos\left(\frac{2l\pi}{n}\right)\right) $$ where $k\in\{0,1,\ldots,m-1\}$ and $l\in\{0,1,\ldots,n-1\}$.